第三章习题讲评(3)LEABX，[BX+20H][SI]MOVAX，[BX+2]解：BX=BX+20H+SI=0056HEA=BX+2=58HPA=DS*16+EA=91D0H+58H=9228HAX=(09228H)=1E40H(4)LDSSI,[BX][DI]MOV[SI]，BX解：EA=BX+DI=56H,PA=DS*16+EA=91D0H+56H=9226HSI=(09226H)=00F6H,DS=(09228H)=1E40HEA=SI=00F6H,PA=DS*16+EA=1E400H+00F6H=1E4F6H(1E4F6H)=BX=0024H(5)XCHGCX，[BX+32H]XCHG[BX+20H][SI]，AX解：EA=BX+32H=56H,PA=DS*16+EA=91D0H+56H=9226H(09226H)=CX=5678H,CX=(09226H)=00F6HEA=BX+20H+SI=56H,PA=DS*16+EA=91D0H+56H=9226HAX=(09226H)=5678H,(09226H)=AX=1234H3.2设DS=1000H，SS=2000H，AX=1A2BH，BX=1200H，CX=339AH，BP=1200H，SP=1350H，SI=1354H，(11350H)=0A5H，(11351H)=3CH，(11352H)=0FFH，(11353H)=26H，(11354H)=52H，(11355H)=OE7H，(126A4H)=9DH，(126A5H)=16H，(21350H)=88H，(21351H)=51H下列各指令都在此环境下执行，在下列各小题的空格中填入相应各指令的执行结果。(1)MOVAX，1352HAX=解：AX=1352H(4)MOVAX，0150H[BP]AX=____解：EA=BP＋0150H＝1350HPA=SS*16+EA=20000H+1350H=21350HAX=5188H(11354H)=0ECH,(11355H)=1AHCF=1,ZF=0,PF(低八位奇偶校验)：0ECH=11101100BPF=0SF(最高位状态)，1H=0001BSF=0OF(溢出标志)0E752H＝1110011101010010B339AH=11001110011010B1110011101010010+1100111001101010001101011101100(7)SUBBH，0150H[BX][SI]BH=____，SF=____，ZF=____，PF=____，CF=____，0F=____解：EA=0150H+BX+SI=26A4H；PA=DS*16+EA=10000H+26A4H=126A4H；(126A4H)=9DH,BH=12HBH=75H,SF=0,ZF=0,PF=0,CF=1,OF=0(10)SARBYTEPTR0150H[BX]，1(11350H)=____，CF=____，OF=____解：EA=BX+0150H=1350HPA＝DS*16+EA=11350H,(11350H)=0A5H=10100101B11010010B=0D2H,CF=1,OF=0(当移位数为1是，最高位不变则OF=0)(11)SALBYTEPTR0150H[BX]，1(11350H)=____，CF=____，OF=____解：EA=BX+0150H=1350H,PA=DS*16+EA=11350,(11350H)=0A5H=10100101B01001010B=4AH,CF=1,OF=13．4阅读下列各小题的指令序列，在后面空格中填入该指令序列的执行结果。(1)MOVBL，85HMOVAL，17HADDAL，BLDAAAL=____，BL=____，CF=____解：17H+85H＝9CHALDAA压缩的BCD码加法十进制调整指令。(AL的低4位>9或AF=1，ALAL+06H,AF1;AF是辅助进位标志用以标志D3向D4的进位AL的高4位>9或CF=1，ALAL+60H,CF1;）AL=9CH+06H=0A2HAL=0A2H+60H=02H,BL=85HCF=1(2)MOVAX，BX；NOTAX；ADDAX，BX；lNCAXAX=____，CF=____解：lNC不影响标志位AX=0000H，CF=0(3)MOVAX，0FF60H；STC；MOVDX，96XORDH，0FFH；SBBAXDXAX=____，CF=____解：XOR命令会使CF0,OF096=60H，AX=0000H